Let the three sides of a triangle ABC be given by the vectors 2i - j + k, i - 3j - 5k and 3i - 4j -

Q: Wait, why do we place vertex A at the origin? Can't we just use the general vector formula for the centroid?

You absolutely can use the general vector formula! However, in competitive exams like JEE Advanced, time is your most precious currency. Placing a vertex at the origin $(0,0,0)$ reduces the algebraic complexity significantly. It turns position vectors into simple coordinates, making the subtraction in the distance formula much less prone to sign errors. It is a strategic choice, not a mathematical necessity.

Q: Does the order of the vectors matter when I assign them to the sides of the triangle?

Yes, it does! Because we identified that $\vec{v_1} + \vec{v_2} = \vec{v_3}$, we know that $\vec{AB} + \vec{BC} = \vec{AC}$. This dictates the orientation. If you swap the vectors, you might end up with a different triangle configuration. Always verify the vector addition loop before assigning them to specific sides.

Q: Is there a shortcut formula for the sum of squared distances to the centroid?

There is a beautiful theorem in geometry: for any triangle with sides $a, b, c$, the sum of the squares of the distances from the vertices to the centroid is given by $\frac{1}{3}(a^2 + b^2 + c^2)$. You could calculate the lengths of the sides, square them, and use this formula. However, calculating the coordinates of $G$ and finding the distances directly is often safer if you aren't 100% confident in recalling the theorem under pressure.

Question

Let the three sides of a triangle $ABC$ be given by the vectors $2\hat{i} - \hat{j} + \hat{k}, \hat{i} - 3\hat{j} - 5\hat{k}$ and $3\hat{i} - 4\hat{j} - 4\hat{k}$. Let $G$ be the centroid of the triangle $ABC$. Then $6\left(|\vec{AG}|^2 + |\vec{BG}|^2 + |\vec{CG}|^2\right)$ is equal to ________

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are not just solving a problem; we are uncovering the hidden symmetry of a triangle in 3D space.

When you first look at these three vectors, $\vec{v_1} = 2\hat{i} - \hat{j} + \hat{k}$, $\vec{v_2} = \hat{i} - 3\hat{j} - 5\hat{k}$, and $\vec{v_3} = 3\hat{i} - 4\hat{j} - 4\hat{k}$, you might feel the urge to jump straight into calculations. But pause. In JEE Advanced, the first step is always observation.

Look at the components. If you add $\vec{v_1}$ and $\vec{v_2}$, you get $3\hat{i} - 4\hat{j} - 4\hat{k}$, which is exactly $\vec{v_3}$. This is the **Triangle Law of Vector Addition** in action: $\vec{AB} + \vec{BC} = \vec{AC}$. This realization is our key to the kingdom.

### The Power of the Origin

Now, let us simplify our world. We have a triangle floating in 3D space. To make our lives easier, let us anchor it.

By placing vertex $A$ at the origin $(0,0,0)$, we effectively turn the position vectors of $B$ and $C$ into their coordinates. Since $\vec{AB} = \vec{v_1}$, vertex $B$ is simply $(2, -1, 1)$. Since $\vec{AC} = \vec{v_3}$, vertex $C$ is $(3, -4, -4)$.

Suddenly, the terrifying 3D geometry has collapsed into simple arithmetic. We are no longer fighting vectors; we are just playing with coordinates.

### The Centroid's Grace

The centroid $G$ is the center of mass of the triangle, the point of perfect balance. The formula is elegant and intuitive: it is the arithmetic mean of the vertices.

$$G = \left( \frac{x_A+x_B+x_C}{3}, \frac{y_A+y_B+y_C}{3}, \frac{z_A+z_B+z_C}{3} ight)$$

Plugging in our values, we get:

$$G = \left( \frac{0+2+3}{3}, \frac{0-1-4}{3}, \frac{0+1-4}{3} ight) = \left( \frac{5}{3}, -\frac{5}{3}, -1 ight)$$

This point $G$ is the heart of our triangle.

### The Final Tally

Now, we calculate the squared distances. We need $|\vec{AG}|^2$, $|\vec{BG}|^2$, and $|\vec{CG}|^2$.

For $|\vec{AG}|^2$, since $A$ is the origin, it is just the sum of the squares of the coordinates of $G$:

$$|\vec{AG}|^2 = \left( \frac{5}{3} ight)^2 + \left( -\frac{5}{3} ight)^2 + (-1)^2 = \frac{25}{9} + \frac{25}{9} + 1 = \frac{59}{9}$$

For $|\vec{BG}|^2$, we find the distance between $B(2,-1,1)$ and $G(\frac{5}{3}, -\frac{5}{3}, -1)$. The differences are $(-\frac{1}{3}, -\frac{2}{3}, -2)$. Squaring these gives:

$$|\vec{BG}|^2 = \frac{1}{9} + \frac{4}{9} + 4 = \frac{41}{9}$$

Finally, for $|\vec{CG}|^2$, the distance between $C(3,-4,-4)$ and $G$ yields $(-\frac{4}{3}, \frac{7}{3}, 3)$. Squaring these gives:

$$|\vec{CG}|^2 = \frac{16}{9} + \frac{49}{9} + 9 = \frac{146}{9}$$

Summing these up, we get:

$$\frac{59+41+146}{9} = \frac{246}{9}$$

The question asks for $6 	imes (	ext{Sum})$. So:

$$6 	imes \frac{246}{9} = 2 	imes \frac{246}{3} = 2 	imes 82 = \mathbf{164}$$

You have done it. You navigated the vectors, anchored the geometry, and executed the algebra with precision. This is the essence of JEE mastery.

Enter Numerical Value:

Visualized Solution

$G i v e nV ec t or s$

$T r ian g l e L a w o f V ec t or A dd i t i o n$

$A ss i g nin g V ec t or s t o S i d es$

$S e tt in g t h e O r i g in$

$C oor d ina t eso f V er t e x B$

$C oor d ina t eso f V er t e x C$

$C e n t r o i d F or m u l a$

$C a l c u l a t in g C e n t r o i d C oor d ina t es$

$C a l c u l a t in g ∣ A G ∣^{2}$

$C a l c u l a t in g ∣ B G ∣^{2}$

$C a l c u l a t in g ∣ C G ∣^{2}$

$S u m o f S q u a r es$

$F ina l C a l c u l a t i o n$

The Sigma Insight: Addition of Vectors

Analyzing the Setup

The Power of the Origin

The Centroid's Grace

The Final Tally

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Enter Numerical Value:

Visualized Solution

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